2. A particle moves along the curve \[ \begin{array}{l}x=2 t^{2}, \quad y=t^{2}-4 t \quad \text { and } \quad z=3 t-5 \\ \text { where } t \text { is the time. Find the components of its velocity and acceleration at time } t=1 \text {, in the } \\ \text { direction } \hat{i}-3 \hat{j}+2 \hat{k} \text {. } \\ \text { (Nagpur, Summer 200l) Ans. } \frac{8 \sqrt{14}}{7},-\frac{\sqrt{14}}{7}\end{array} \]

To solve the problem of finding the components of the velocity and acceleration vectors in the direction of the vector \(\mathbf{d} = \hat{i} - 3\hat{j} + 2\hat{k}\) at time \(t = 1\), we'll follow these steps: 1. **Determine the Position Vector \(\mathbf{r}(t)\)** Given: \[ x = 2t^2, \quad y = t^2 - 4t, \quad z = 3t - 5 \] The position vector is: \[ \mathbf{r}(t) = 2t^2\, \hat{i} + (t^2 - 4t)\, \hat{j} + (3t - 5)\, \hat{k} \] 2. **Find the Velocity Vector \(\mathbf{v}(t)\)** The velocity vector is the first derivative of the position vector with respect to time \(t\): \[ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = 4t\, \hat{i} + (2t - 4)\, \hat{j} + 3\, \hat{k} \] At \(t = 1\): \[ \mathbf{v}(1) = 4(1)\, \hat{i} + (2(1) - 4)\, \hat{j} + 3\, \hat{k} = 4\, \hat{i} - 2\, \hat{j} + 3\, \hat{k} \] 3. **Find the Acceleration Vector \(\mathbf{a}(t)\)** The acceleration vector is the derivative of the velocity vector with respect to time \(t\): \[ \mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = 4\, \hat{i} + 2\, \hat{j} + 0\, \hat{k} \] At \(t = 1\): \[ \mathbf{a}(1) = 4\, \hat{i} + 2\, \hat{j} + 0\, \hat{k} \] 4. **Determine the Unit Vector in the Direction of \(\mathbf{d}\)** Given \(\mathbf{d} = \hat{i} - 3\hat{j} + 2\hat{k}\), the magnitude of \(\mathbf{d}\) is: \[ |\mathbf{d}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] Therefore, the unit vector \(\mathbf{\hat{d}}\) in the direction of \(\mathbf{d}\) is: \[ \mathbf{\hat{d}} = \frac{1}{\sqrt{14}}\, \hat{i} - \frac{3}{\sqrt{14}}\, \hat{j} + \frac{2}{\sqrt{14}}\, \hat{k} \] 5. **Compute the Components of Velocity and Acceleration in the Direction of \(\mathbf{d}\)** The component of a vector \(\mathbf{A}\) in the direction of \(\mathbf{\hat{d}}\) is given by the dot product: \[ \text{Component} = \mathbf{A} \cdot \mathbf{\hat{d}} \] - **Velocity Component:** \[ \mathbf{v}(1) \cdot \mathbf{\hat{d}} = (4)(1) + (-2)(-3) + (3)(2) = 4 + 6 + 6 = 16 \] \[ \text{Velocity Component} = \frac{16}{\sqrt{14}} = \frac{8\sqrt{14}}{7} \] - **Acceleration Component:** \[ \mathbf{a}(1) \cdot \mathbf{\hat{d}} = (4)(1) + (2)(-3) + (0)(2) = 4 - 6 + 0 = -2 \] \[ \text{Acceleration Component} = \frac{-2}{\sqrt{14}} = -\frac{\sqrt{14}}{7} \] **Final Answer:** - **Component of Velocity:** \(\dfrac{8\, \sqrt{14}}{7}\) - **Component of Acceleration:** \(-\dfrac{\sqrt{14}}{7}\)